Is the series of function $\sum_{n = 0}^{\infty} \frac{x^n} {n}$ uniformly convergent on $(0, 1)$?

The original question asked for which $b > 0$, the series of function $\sum_{n = 0}^{\infty} \frac{x^n} {n}$ is uniformly convergent on $(0, b)$.

When $0 < b < 1$, we know $\sum_{n = 0}^{\infty} \frac{b^n} {n}$ converges by root test. So by Weierstrass's M-test, $\sum_{n = 0}^{\infty} \frac{x^n} {n}$ is uniformly convergent on $(0, b)$.

On the other hand, when $b > 1$, $\sum_{n = 0}^{\infty} \frac{x^n} {n}$ is not even pointwisely convergent at $1$ since the harmonic series diverges.

But I am stuck on $b = 1$, how should I proceed?

EDIT-1:

The following is from Bartle's Introduction to Real Analysis:

9.4.6 Weierstrass M-Test Let $(M_n)$ be a sequence of positive real numbers such that $|f_n(x)| \le M_n$ for $x \in D, n \in \mathbb{N}$. If the series $\sum M_n$ is convergent, then $\sum f_n$ is uniformly convergent on $D$.

• See the first answer here. – David Mitra Apr 2 '17 at 12:08
• @DavidMitra Thank you, so I need to use the Cauchy criterion to show that $\sum_{n = 0}^{\infty} \frac{x^n} {n}$ is not uniformly convergent on $(0, 1)$. – Alex Vong Apr 2 '17 at 12:42
• @DavidMitra At that link you say in a comment that the partial sums are not bounded, which I don't think is what you meant to say. – zhw. Apr 3 '17 at 4:32